Pythagoras and custom angles?

Title is messed up, It should really be trigonometry..

For example I have a right angled triangle, no matter what.

So you would have opposite, adjacent and hypotenuse..


Now imagine, the angle (A) is in the top right, between Hyp and Opp..

(Question 1): With only the Angle and Opposite for known's is it possible to figure out the meeting point of the Hypotenuse and Adjacent?

THEN, to make things a little trickier.. After the angle of incident you would have an angle of reflection, giving a second leg to the Hypotenuse and thus creating another right angled triangle in reflection to this one.

(Question 2): Is it then possible, to know after finding the angle of incident and how far along the Adjacent that would be. To then given off a particular "Length" of the Hypotenuse, that goes beyond the angle of incident resulting in an angle of reflection to then figure out the Adjacent and Opposite of the second triangle ( Angle of reflection )..?

Just curious as to the formulae to such an equation..

EDIT:

Answers Required;

Incident Length of Adjacent?

Opposite and Adjacent Lengths of a hypotenuse that exceeds the meeting point based on the initial angle?

Also, if anyone has a brain much larger than that, you could factor in Snell's Law to give an angle of refraction rather than reflection if you see fit !

 

Isn't this just SOHCAHTOA?

 

https://www.youtube.com/watch?v=F21S9Wpi0y8 helpes to watch this dood

 

That's correct, it's trigonometry, I've already figured this out. As well as adding beam paths either side dependant on the beam divergence. I was just curious as to whether anyone else fancied trying it.